Question #272317

Q5. A 5 kg of ice cube at -10°C is mixed with 0.1 kg of water at 80 °C. There is no heat





loss to the surrounding. The specific heat capacity of ice is 2.22 kJK-1kg ? The specific





heat capacity of water is 4.187 kJK-kg“The specific latent heat of fusion of ice is





333 kJ kg-1





(a) What is the final physical state of the mixture? (Gas, Liquid or Solid?)





Justify your answer using suitable calculations





(b) What is the final temperature of the mixture?

1
Expert's answer
2021-11-29T11:47:50-0500

a)

Qitot=cimi(0ti)=111 kJ,Q_{i_{tot}}=c_im_i(0-t_i)=111~kJ,

Qwtot=cwmw(tw0)=33.5 kJ,Q_{w_{tot}}=c_wm_w(t_w-0)=33.5~kJ,

Qitot>Qwtot,Q_{i_{tot}}>Q_{w_{tot}}, it will be an ice,

b)

Qi=Qwtot+Qwλ+Qwi,Q_i=Q_{w_{tot}}+Q_{w_{\lambda}}+Q_{w_i},

cimi(tti)=cwmw(tw0)+λmw+cimw(0t),c_im_i(t-t_i)=c_wm_w(t_w-0)+\lambda m_w+c_im_w(0-t),

t=mw(λ+cwtw)+cimitici(mi+mw)=3.9°C.t=\frac{m_w(\lambda+c_wt_w)+c_im_it_i}{c_i(m_i+m_w)}=-3.9°C.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS